For example, given the following code
class A {
public:
double operator()(double foo) {
return foo;
}
};
class B {
public:
double operator()(double foo, int bar) {
return foo + bar;
}
};
I want to write two versions of fun
, one that works with objects with A's signature and another one that works with objects with B's signature:
template <typename F, typename T>
T fun(F f, T t) {
return f(t);
}
template <typename F, typename T>
T fun(F f, T t) {
return f(t, 2);
}
And I expect this behavior
A a();
B b();
fun(a, 4.0); // I want this to be 4.0
fun(b, 4.0); // I want this to be 6.0
Of course the previous example throws a template redefinition error at compile time.
If B is a function instead, I can rewrite fun
to be something like this:
template <typename T>
T fun(T (f)(T, int), T t) {
return f(t, 2);
}
But I want fun
to work with both, functions and callable objects. Using std::bind
or std::function
maybe would solve the problem, but I'm using C++98 and those were introduced in C++11.
Here's a solution modified from this question to accommodate void-returning functions. The solution is simply to use sizeof(possibly-void-expression, 1)
.
#include <cstdlib>
#include <iostream>
// like std::declval in c++11
template <typename T>
T& decl_val();
// just use the type and ignore the value.
template <std::size_t, typename T = void>
struct ignore_value {typedef T type;};
// This is basic expression-based SFINAE.
// If the expression inside sizeof() is invalid, substitution fails.
// The expression, when valid, is always of type int,
// thanks to the comma operator.
// The expression is valid if an F is callable with specified parameters.
template <class F>
typename ignore_value<sizeof(decl_val<F>()(1),1), void>::type
call(F f)
{
f(1);
}
// Same, with different parameters passed to an F.
template <class F>
typename ignore_value<sizeof(decl_val<F>()(1,1),1), void>::type
call(F f)
{
f(1, 2);
}
void func1(int) { std::cout << "func1\n"; }
void func2(int,int) { std::cout << "func2\n"; }
struct A
{
void operator()(int){ std::cout << "A\n"; }
};
struct B
{
void operator()(int, int){ std::cout << "B\n"; }
};
struct C
{
void operator()(int){ std::cout << "C1\n"; }
void operator()(int, int){ std::cout << "C2\n"; }
};
int main()
{
call(func1);
call(func2);
call(A());
call(B());
// call(C()); // ambiguous
}
Checked with gcc and clang in c++98 mode.